alexandria/documents/by-course/pstat-120a/hw1/main.typ

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#import "@youwen/zen:0.1.0": *
#import "@preview/ctheorems:1.1.3": *
#show: zen.with(
title: "Homework 1",
author: "Youwen Wu",
date: "Winter 2025",
)
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#set enum(spacing: 2em)
+ #[
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#set enum(numbering: "a)", spacing: 2em)
+ #[
We know that $B$ and $B'$ are disjoint. That is, $B sect B' =
emptyset$. Additionally,
$
E = (A sect B) subset B \
F = (A sect B') subset B' \
$
Then we note
$
forall x in E, x in B, x in.not B' \
forall y in F, y in B, y in.not B'
$
So clearly $E$ and $F$ have no common elements, and
$
E sect F = emptyset
$
]
+ #[
$
E union F &= (A sect B) union (A sect B') \
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&= (A union A) sect (B union B') \
&= A sect Omega \
&= A
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$
]
]
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+ #[
#set enum(numbering: "a)", spacing: 2em)
+ ${15, 25, 35, 45, 51, 53, 55, 57, 59, 65, 75, 85, 95 }$
+ ${50, 52, 56, 58}$
+ $emptyset$
]
+ #[
#set enum(numbering: "a)", spacing: 2em)
+ The sample space is every value of the die (1-6) paired with heads and paired with tails. That is, ${1,2,3,4,5,6} times {H,T}$, with cardinality 12.
+ There are $12^10$ outcomes.
+ If no participants roll a 5, then we omit any outcome in our sample space where the die outcome is 5, leaving us with 10 outcomes of the die and coin. Now we have $10^10$ outcomes. If at least 1 person rolls a 5, then we note that this is simply the complement of the previous result. So we have $12^10 - 10^10$ outcomes total.
]
+ #[
#set enum(numbering: "a)", spacing: 2em)
+ #[
The sample space can be represented as a 6-tuple where the position 1-6
represents balls numbered 1-6, and the value represents the square it
was sent to. So it's
$ {{x_1,x_2,x_3,x_4,x_5,x_6} : x_i in {1,2,3,4}}, i = 1,...6 $
]
+ #[
When the balls are indistinguishable, we can instead represent it as
4-tuples where the position represents the 1st, 2nd, 3rd, or 4th square,
and the value represents how many balls landed. Additionally the sum of all
the elements must be 6.
$ {{x_1, x_2, x_3, x_4} : x_i >= 0, i = 1,...,6 sum_(j=1)^4 x_j = 6} $
]
]